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Hi, Ofcourse the above method by Bunuel is the right and best method, But what it means logically..

is \(|a-b| > |a+b|\)?.. We have a number line in which a and b and -b are marked.. the MOD asks us whether a is farther from b than -b..

two cases-

A) if both are same sign, a will be closer to b than -b, since midpoint of b and -b will be 0.. ans will be NO B) if both are opposite sign, a will be closer to -b and ans will be YES..

1) \(ab < 0\) Both a and b are of opposite sign .. ans YES Suff

2) \(a > b\) Nothing .. both can be same sign or opposite sign .. the above TWO cases still remain.. Insuff
_________________

Hi, Ofcourse the above method by Bunuel is the right and best method, But what it means logically..

is \(|a-b| > |a+b|\)?.. We have a number line in which a and b and -b are marked.. the MOD asks us whether a is farther from b than -b..

two cases-

A) if both are same sign, a will be closer to b than -b, since midpoint of b and -b will be 0.. ans will be NO B) if both are opposite sign, a will be closer to -b and ans will be YES..

1) \(ab < 0\) Both a and b are of opposite sign .. ans YES Suff

2) \(a > b\) Nothing .. both can be same sign or opposite sign .. the above TWO cases still remain.. Insuff

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Hi, Ofcourse the above method by Bunuel is the right and best method, But what it means logically..

is \(|a-b| > |a+b|\)?.. We have a number line in which a and b and -b are marked.. the MOD asks us whether a is farther from b than -b..

two cases-

A) if both are same sign, a will be closer to b than -b, since midpoint of b and -b will be 0.. ans will be NO B) if both are opposite sign, a will be closer to -b and ans will be YES..

1) \(ab < 0\) Both a and b are of opposite sign .. ans YES Suff

2) \(a > b\) Nothing .. both can be same sign or opposite sign .. the above TWO cases still remain.. Insuff

Hi chetan2u,

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Thanks in advance

Take a look at my previous answer. You must analyze four different situations because the two sides of the equation are not independent.

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Thanks in advance

Hi,

the problem in your solution is opening of MOD.. yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it..

Do not waste time in taking four scenarios, but square both sides, it will siffice..

\(|a-b|>|a+b|..\) square both sides.. \((a-b)^2 > (a+b)^2................... a^2+b^2-2ab>a^2+b^2+2ab............ 4ab<0....\).... so we should find out if ab<0

and statement I tells us that ab<0 .. so suff
_________________

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Thanks in advance

Hi,

the problem in your solution is opening of MOD.. yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it..

Do not waste time in taking four scenarios, but square both sides, it will siffice..

\(|a-b|>|a+b|..\) square both sides.. \((a-b)^2 > (a+b)^2................... a^2+b^2-2ab>a^2+b^2+2ab............ 4ab<0....\).... so we should find out if ab<0

and statement I tells us that ab<0 .. so suff

Thanks Z10201Z for your help

Hi chetan2u,

Thanks you for your quick response. I certainly will follow the rule in case of two modulus with inequality.

Aster reading your post about 'Absolute modulus : A better understanding', I have two questions:

1- I want to take further to solve a confusion. I have seen your way solving in one problem when two modulus in equality: |X|=|Y|

X=Y or X=-Y or -X=-Y (which is same as X=Y) or -X=Y (Which is the same as X=-Y). However, the previous technique is valid only for Equity and invalid for inequality. Am I right? I think this my error.

2- if the question is |a-b|>a+b ............what the best to solve it? In the case, I'm curious to know how to open modulus just to get to know.

I hope one day you can make another post for absolute modulus under inequality.

I tried other approach by opening the modulus but it seems there is an error.

is |a−b|>|a+b|?

case 1:

a-b > a+b……… 2b<0……..so b<0.. B is negative

case 2:

a-b < -a-b …….. 2a<0……..so a< 0 ..a is negative

Statement 1:

ab<0

According to my analysis above, it should be insufficient because both a & b should negative.

Statement 2:

a>b It should be insufficient

Can you help with identifying the problem with above solution?

Thanks in advance

Hi,

the problem in your solution is opening of MOD.. yes, you can save a lot of time in opening of MOD, if both sides are +ive but do not open it straightway, SQUARE it..

Do not waste time in taking four scenarios, but square both sides, it will siffice..

\(|a-b|>|a+b|..\) square both sides.. \((a-b)^2 > (a+b)^2................... a^2+b^2-2ab>a^2+b^2+2ab............ 4ab<0....\).... so we should find out if ab<0

and statement I tells us that ab<0 .. so suff

Thanks Z10201Z for your help

Hi chetan2u,

Thanks you for your quick response. I certainly will follow the rule in case of two modulus with inequality.

Aster reading your post about 'Absolute modulus : A better understanding', I have two questions:

1- I want to take further to solve a confusion. I have seen your way solving in one problem when two modulus in equality: |X|=|Y|

X=Y or X=-Y or -X=-Y (which is same as X=Y) or -X=Y (Which is the same as X=-Y). However, the previous technique is valid only for Equity and invalid for inequality. Am I right? I think this my error.

2- if the question is |a-b|>a+b ............what the best to solve it? In the case, I'm curious to know how to open modulus just to get to know.

I hope one day you can make another post for absolute modulus under inequality.

Thanks in advance for your help. I appreciate it.

hi lets solve |a-b|>a+b.. this can be straight .. 1) a-b>a+b......2b<0..... b<0.. 2) -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..

Or Critical points..

a is 0, when a is b, so b is our critical point.. this gives us two ranges a<b or a>=b.. 1) a<b... then a-b will be -ive.. so -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0.. 2) a>=b.. then a-b will be +ive or 0.. so a-b>a+b......2b<0..... b<0..

Best METHOD for any MOD irrespective of = or inequality, follow "critical point" method
_________________

hi lets solve |a-b|>a+b.. this can be straight .. 1) a-b>a+b......2b<0..... b<0.. 2) -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0..

Or Critical points..

a is 0, when a is b, so b is our critical point.. this gives us two ranges a<b or a>=b.. 1) a<b... then a-b will be -ive.. so -(a-b)>a+b......-a+b>a+b.....2a<0..... a<0.. 2) a>=b.. then a-b will be +ive or 0.. so a-b>a+b......2b<0..... b<0..

Best METHOD for any MOD irrespective of = or inequality, follow "critical point" method

Thanks a lot for your reply.

I think there is typo. it should: a-b=0, when a is b

Could some one please help me with this question with a valid explanation. Thanks , Shiv.

According to statement 1. a*b is negative. So, either a or b is negative. Both of them can't be negative. by question a*b doesn't equals to zero. take testing numbers. a = 1 or -1 b = -1 or 1 if a = 1 and b =-1 then; |1 -(-1)| > |1 -1| or, |1+1| > |0| or, 2>0. true.

also, check for a= -1 and b =1 |-1-(1)| > |-1+1| or, |-1-1| > |0| or, 2>0. true. so Statement 1 is sufficient.

According to statement 2, a>b. take testing points. a=-1 or 2 b= -2 or -1 if a= -1 and b = -2 then, |-1 - (-2)| > |-1+(-2)| or, |-1+2|> |-1-2| or, |1|>|-3| or 1>3 false.

If a= 2 and b = -1 |2-(-1)|>|2-1| or,|3|>|1| or, 3>1 true. So, statement 2 is not sufficient to answer the question. Statement 2 gives either true or false. Whereas, Statement 1 gives true only.

So, answer A.

I was waiting for an easy solution and Vyshak gives that.

You can think of absolute value as distance from zero. So we need enough info to definitely determine whether a - b is further from zero than the sum a + b.

Statement one tells us that a and b have different signs (that's the only way that the product will be negative).

For the difference a - b, this means that the numbers end up 'working together' to get away from zero; if you take a positive number away from a negative number, you go further down, further away from zero (e.g. -10 - 2 = -12). If you take a negative number away from a positive number you go further up (e.g. 10 - - 2 = 12) Takeaway: The difference of numbers with opposite signs is always further from zero than either of the original numbers.

For the sum a + b, statement one means that the numbers end up 'working in opposite directions.' That is, when we add numbers of opposite signs, one cancels out part of the other's distance to zero. Think of 10 + - 2 and -2 + 10. In both cases we end up closer to zero than the furthest term was to begin with. Takeaway: the sum of two numbers with opposite signs will be closer to zero than the original number of greatest magnitude (the term that was furthest from zero to begin with).

So if we know two numbers have opposite signs, we know the difference of those numbers ends up further from zero than the sum. Sufficient.

statement 2 doesn't tell us anything about the signs, so we might be suspicious right from the start. And sure enough, if all we know is that a is to the right of b, then we might have a couple of positives, (such as a = 3, and b = 2) in which case the sum is further from zero than the difference. But we could also have a positive and a negative (such as a = 3, and b = - 2), in which case the difference is further from zero than the sum. Insufficient.

However, I would like to know how this problem will be solved algebrically. Can you please explain the solution by opening the modulus on both the side ?

There will be following two condition if we open modulus:

a-b > a+b gives, b <0

and

a-b > -(a+b) gives , a>0

What would be the next steps now ? How to interpret. Please explain.

On the left side we have sum of two positive integers so we can pull off our modulus. On the right side we have difference o two positive integers which will always be lesser than their sum.

a<0, b>0. Same approach. a=-x (x some positive integer). -x<0, x>0

|-x - b| > |-x + b|

| - (x +b)| > |b - x|

x + b > |b - x|

Again we have sum of two positive integers on the left and their diffrence on th right. Sufficient.

2) a > b

We have three possible scenarious.

a) a>0, b>0 b) a<0, b<0 c) a>0, b<0

a) To eveade confusion it would be impler to use particular numbers here.

a = 30, b = 10

|30 - 10| ... |30 + 10| ----> 20 < 40

b) a = -10 b = -30

|-10 - (-30)| ... | -10 - 30|

| -10 + 30| ... | - (10 + 30)|

20 < 40

c) In c we have same case as in (1)

a=10, b=-30

|10 - (-30)| ... |10 - 30|

40 > 20 Insufficient

Answer A

I hope my approach wasn't too vague. May be our experts will give you more details.